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Chapter 0: Problem 64

find the solutions of the equation in \([0,2 \pi)\). $$ \csc ^{2} x-\cot x-1=0 $$

Short Answer

Expert verified
The solutions of the equation \(\csc^2 x - \cot x - 1 = 0\) on the interval \([0, 2\pi)\) are \(x = \frac{\pi}{2}\) and \(x = \frac{3 \pi}{2}\).

Step by step solution

01

Rewrite the equation in terms of sine and cosine

Recall that the cosecant function is defined as \(\csc x = \frac{1}{\sin x}\). And the cotangent function is defined as \(\cot x = \frac{\cos x}{\sin x}\). Rewrite the equation as follows: \[\left(\frac{1}{\sin^2 x}\right) - \left(\frac{\cos x}{\sin x}\right) - 1 = 0\]
02

Simplify the equation

To simplify the equation, we can start by multiplying both sides with a common denominator, which in this case is \(\sin^2 x\): \[\sin^2 x\left(\frac{1}{\sin^2 x}\right) - \sin^2 x\left(\frac{\cos x}{\sin x}\right) - \sin^2 x(1) = 0\] This simplifies to: \[1 - \cos x \sin x - \sin^2 x = 0\]
03

Express in terms of sine

Now, let's express \(\cos x\) in terms of \(\sin x\) using the identity \(\cos^2 x + \sin^2 x = 1\): \[\cos^2 x = 1 - \sin^2 x\] So, we can rewrite the equation as: \[1 - \sqrt{1 - \sin^2 x} \sin x - \sin^2 x = 0\]
04

Solve the resulting equation

Let \(u = \sin x\), and the equation becomes a quadratic equation: \[1 - \sqrt{1 - u^2} u - u^2 = 0\] Now, to solve this equation, we can first isolate the radical term: \[\sqrt{1 - u^2} u = 1 - u^2\] Square both sides to eliminate the radical: \[(1 - u^2) u^2 = (1 - u^2)^2\] Expand and simplify: \[u^4 - 2u^2 + 1 = u^4 - 2u^2 + u^2\] Subtract \(u^4 - 2u^2\) from both sides: \[u^2 = 1\] Taking the square root of both sides: \[u = \pm 1\] So, the solutions are: \[\sin x = \pm 1\]
05

Find the values of x

On the given interval \(0 \leq x < 2 \pi\), we have: For \(\sin x = 1\): \[x = \frac{\pi}{2}\] For \(\sin x = -1\): \[x = \frac{3 \pi}{2}\] The solutions for the given equation are: \(x = \frac{\pi}{2}, \frac{3 \pi}{2}\) on the interval [0, \(2\pi\)).

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Most popular questions from this chapter

Plot the graph of the function \(f\) in an appropriate viewing window. (Note: The answer is not unique.) $$ f(x)=\frac{1}{2+\cos x} $$

Suppose that \(f\) is a one-to-one function such that \(f(3)=7\) Find \(f\left[f^{-1}(7)\right]\).

Find the inverse of \(f .\) Then use a graphing utility to plot the graphs of \(f\) and \(f^{-1}\) using the same viewing window. $$ f(x)=\frac{x}{x^{2}+1}, \quad-\frac{1}{2} \leq x \leq \frac{1}{2} $$

a. Plot the graph of \(f(x)=\sqrt{x} \sqrt{x-1}\) using the viewing window \([-5,5] \times[-5,5]\). b. Plot the graph of \(g(x)=\sqrt{x(x-1)}\) using the viewing window \([-5,5] \times[-5,5]\). c. In what interval are the functions \(f\) and \(g\) identical? d. Verify your observation in part (c) analytically.

Find \(f^{-1}(a)\) for the function \(f\) and the real number \(a\). $$ f(x)=2+\tan \left(\frac{\pi x}{2}\right), \quad-1
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